NCERT Solutions for Exercise 13.2 Class 9 Maths Chapter 13 - Surface Area and Volumes

# NCERT Solutions for Exercise 13.2 Class 9 Maths Chapter 13 - Surface Area and Volumes

Edited By Vishal kumar | Updated on Oct 13, 2023 09:11 AM IST

## NCERT Solutions for Class 9 Maths Exercise 13.2 Chapter 13 Surface Areas And Volumes- Download Free PDF

NCERT Solutions for Class 9 Maths Exercise 13.2 Chapter 13 Surface Areas And Volumes- NCERT Solutions for Class 9 Maths Exercise 13.2 deals with the concept of the right circular cylinder and its surface areas. A right circular cylinder is a cylinder with a closed circular surface, two parallel bases on both ends and elements that are perpendicular to the base. Three pieces make up the right circular cylinder. They are

• Top circular base

• Curved lateral face

• Bottom circular face

In exercise 13.2 Class 9 Maths, A combination of two circles plus a rectangle forms the right circular cylinder. The surface area of a right circular cylinder is the area covered by the surface of the right circular cylinder. Surface area can be divided into two categories. They are

• Lateral Surface Area

• Total Surface Area

The curved surface area of the right circular cylinder, also known as the lateral surface area of the right circular cylinder, is the area covered by the curved surface of the cylinder. The total surface area of the right circular cylinder is the area occupied by the complete cylinder. NCERT solutions for Class 9 Maths chapter 13 exercise 13.2 consists of 11 questions which are direct formula based questions. The concepts related to the surface area and volumes are well explained in this NCERT syllabus 9th class maths exercise 13.2 answers . Along with NCERT book Class 9 Maths chapter 13 exercise 13.2 the following exercises are also present.

***According to the CBSE Syllabus 2023-24, this chapter is now Chapter 11.

## Access Surface Area and Volumes Class 9 Chapter 13 Exercise: 13.2

Given,

The curved surface area of the cylinder =

And, the height of the cylinder,

We know, Curved surface area of a right circular cylinder =

Therefore, the diameter of the cylinder =

Given,

Height of the cylindrical tank = $h = 1\ m$

Base diameter = $\small d = 140\ cm = 1.4\ m$

We know,

The total surface area of a cylindrical tank = $\small 2\pi r h+2\pi r^2 = 2\pi r(r+h)$

$\small = 2.\frac{22}{7}. \frac{1.4}{2}.(0.7+1) = 2.\frac{22}{7}. (0.7).(1.7)$

$\small = 44\times0.17$

$\small = 7.48\ m^2$

Therefore, square metres of the sheet is $\small 7.48\ m^2$

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder, $h = 77\ cm$

Outer diameter = $r_1 = 4.4\ cm$

Inner diameter = $r_2 = 4\ cm$

Inner curved surface area = $2\pi r_2h$

$\\ = 2\times\frac{22}{7}\times2\times77 \\ = 968\ cm^2$

Therefore, the inner curved surface area of the cylindrical pipe is $968\ cm^2$

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder, $h = 77\ cm$

Outer diameter = $r_1 = 4.4\ cm$

Inner diameter = $r_2 = 4\ cm$

Outer curved surface area = $2\pi r_1h$

$\\ = 2\times\frac{22}{7}\times2.2\times77 \\ = 1064.8\ cm^2$

Therefore, the outer curved surface area of the cylindrical pipe is $1064.8\ cm^2$

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder, $h = 77\ cm$

Outer diameter = $r_1 = 4.4\ cm$

Inner diameter = $r_2 = 4\ cm$

Outer curved surface area = $2\pi r_1h$

Inner curved surface area = $2\pi r_2h$

Area of the circular rings on top and bottom = $2\pi(r_2^2-r_1^2)$

$\therefore$ The total surface area of the pipe = $2\pi r_1h +2\pi r_2h+ 2\pi(r_2^2-r_1^2)$

$\\ = [968 + 1064.8 + 2\pi {(2.2)^2 - (2)^2}]\\ \\ = (2032.8 + 2\times \frac{22}{7}\times 0.84) \\ \\ = (2032.8 + 5.28) \\ = 2038.08\ cm^2$

Therefore, the total surface area of the cylindrical pipe is $2038.08\ cm^2$

Given,

The diameter of the cylindrical roller = $\small d = 84\hspace{1mm}cm$

Length of the cylindrical roller = $\small h = 120\hspace{1mm}cm$

The curved surface area of the roller = $2\pi r h = \pi dh$

$\\ = \frac{22}{7}\times84\times120 \\ \\ = 31680\ cm^2$

$\therefore$ Area of the playground = $Area\ covered\ in\ 1\ rotation \times500$

$\\ = 31680 \times500 \\ = 15840000 cm^2 \\ = 1584\ m^2$

Therefore, the required area of the playground = $1584\ m^2$

Given,

Radius of the cylindrical pillar, r = $\frac{50}{2}\ cm= 25\ cm= 0.25\ m$

Height of the cylinder, h = $3.5\hspace{1mm}m$

We know,

Curved surface area of a cylinder = $2\pi r h$

$\therefore$ Curved surface area of the pillar = $2\times\frac{22}{7}\times 0.25 \times3.5$

$= 1\times22\times 0.25\ m^2$

$= 5.5\ m^2$

Now,

Cost of painting $\small 1\ m^2$ of the pillar = $\small Rs \hspace{1mm} 12.50$

$\therefore$ Cost of painting the curved surface area of the pillar = $\small Rs.\ (12.50\times5.5)$

$\small = Rs.\ 68.75$

Therefore, the cost of painting curved surface area of the pillar is $\small Rs.\ 68.75$

Given, a right circular cylinder

Curved surface area of the cylinder = $\small 4.4\hspace{1mm}m^2$

The radius of the base = $r = 0.7\ m$

Let the height of the cylinder be $h$

We know,

Curved surface area of a cylinder of radius $r$ and height $h$ = $2\pi r h$

$\therefore$ $2\pi r h = 4.4$

$\\ \Rightarrow 2\times\frac{22}{7}\times0.7 \times h = 4.4 \\ \\ \Rightarrow 44\times0.1 \times h = 4.4 \\ \Rightarrow h = 1\ m$

Therefore, the required height of the cylinder is $1\ m$

Given,

The inner diameter of the circular well = $d = \small 3.5\hspace{1mm}m$

Depth of the well = $h = 10\ m$

We know,

The curved surface area of a cylinder = $2\pi rh$

$\therefore$ The curved surface area of the well = $2\times\frac{22}{7}\times\frac{3.5}{2}\times10$

$\\ = 44 \times 0.25 \times 10 \\ = 110\ m^2$

Therefore, the inner curved surface area of the circular well is $110\ m^2$

Given,

The inner diameter of the circular well = $d = \small 3.5\hspace{1mm}m$

Depth of the well = $h = 10\ m$

$\therefore$ The inner curved surface area of the circular well is $110\ m^2$

Now, the cost of plastering the curved surface per $\small m^2$ = Rs. 40

$\therefore$ Cost of plastering the curved surface of $110\ m^2$ = $Rs.\ (110 \times 40) = Rs.\ 4400$

Therefore, the cost of plastering the well is $Rs.\ 4400$

Given,

Length of the cylindrical pipe = $l = \small 28\hspace{1mm}m$

Diameter = $\small d = 5\hspace{1mm}cm = 0.05\ m$

The total radiating surface will be the curved surface of this pipe.

We know,

The curved surface area of a cylindrical pipe of radius $r$ and length $l$ = $2\pi r l$

$\therefore$ CSA of this pipe =

$= 2\times\frac{22}{7}\times\frac{0.05}{2}\times28$

$\\ = 22\times0.05\times4 \\ = 4.4\ m^2$

Therefore, the total radiating surface of the system is $4.4\ m^2$

Given, a closed cylindrical petrol tank.

The diameter of the tank = $\small d = 4.2\hspace{1mm}m$

Height of the tank = $\small h = 4.5\hspace{1mm}m$

We know,

The lateral surface area of a cylinder of radius $\small r$ and height $\small h$ = $\small 2\pi r h$

$\small \therefore$ The lateral surface area of a cylindrical tank = $\small 2\times\frac{22}{7}\times\frac{4.2}{2}\times4.5$

$\small \\ = (44 \times 0.3 \times 4.5) \\ = 59.4\ m^2$

Therefore, the lateral or curved surface area of a closed cylindrical petrol storage tank is $\small 59.4\ m^2$

Given, a closed cylindrical petrol tank.

The diameter of the tank = $\small d = 4.2\hspace{1mm}m$

Height of the tank = $\small h = 4.5\hspace{1mm}m$

Now, Total surface area of the tank = $2\pi r (r + h)$

$\\ = 2 \times \frac{22}{7} \times 2.1 \times (2.1 + 4.5) \\ = (44 \times 0.3 \times 6.6) \\ = 87.12\ m^2$

Now, let $x\ m^2$ of steel sheet be actually used in making the tank

Since $\small \frac{1}{12}$ of steel was wasted, the left $\small \frac{11}{12}$ of the total steel sheet was used to made the tank.
$\small \therefore$ The total surface area of the tank = $\small \frac{11}{12}\times (Total\ area\ of\ steel\ sheet )$
$\small \\ \Rightarrow 87.12= \frac{11}{12}x \\ \Rightarrow x = \frac{87.12\times12}{11} \\ \Rightarrow x = 95.04\ m^2$

Therefore, $\small 95.04\ m^2$ of steel was actually used in making the tank.

The diameter of the base = $d = 20\ cm$

Height of the cylinder = $30\ cm$

The total height of lampshade= $(30+ 2.5 + 2.5) = 35\ cm$

We know,

Curved surface area of a cylinder of radius $r$ and height $h$ = $2\pi r h$

Now, Cloth required for covering the lampshade = Curved surface area of the cylinder

$\\ = 2\times\frac{22}{7} \times10 \times 35 \\ \\ = 22\times10\times10 = 2200\ cm^2$

Therefore, $2200\ cm^2$ cloth will be required for covering the lampshade.

Given, a cylinder with a base.

The radius of the cylinder = $r = 3\ cm$

Height of the cylinder = $h = 10.5\ cm$

We know,

The lateral surface area of a cylinder of radius $r$ and height $h$ = $2\pi r h$

$\therefore$ Area of the cylindrical penholder = Lateral areal + Base area

$= 2\pi r h + \pi r^2 = \pi r (2h+r)$

$= \frac{22}{7}\times3\times[2(10.5)+3]$

$\\ = \frac{22}{7}\times3\times[24] \\ \\ = \frac{1584}{7}\ cm^2$
Area of 35 penholders = $\frac{1584}{7}\times35\ cm^2$

$= 7920\ cm^2$

Therefore, the area of carboard required is $7920\ cm^2$

## More about NCERT Solutions for Class 9 Maths Exercise 13.2 :

The NCERT solutions for Class 9 Maths exercise 13.2 is mainly focused on the surface area of the right circular cylinder. In Exercise 13.2 Class 9 Maths, the lateral surface area of the closed right circular cylinder can be calculated by the product of 2π and the radius and height of the cylinder. CSA= 2πrh . The total surface area of a closed right circular cylinder is the sum of the area of the lateral surface and the area of the two bases.

TSA= CAS + 2(Area of a circle)

= 2πrh+2πr2

TSA=2πr(h+r)

The volume of the right circular cylinder was calculated by the product of the area of a circle and the height of the cylinder. V=πr2h. Surface areas are measured in square units, but the volume of a cube is measured in cubic units. In NCERT solutions for Class 9 Maths Exercise 13.2, the formulas for computing surface areas and volume for the correct circular cylinder are thoroughly covered.

Also Read| Surface Areas And Volumes Class 9 Notes

## Benefits of NCERT Solutions for Class 9 Maths Exercise 13.2 :

• NCERT solutions for Class 9 Maths Exercise 13.2 gives a clear outline of the right circular cylinder in detail so that we get a clear understanding of what it is.

• By answering the NCERT solution for Class 9 Maths chapter 13 exercise 13.2 exercises, we can improve our grades in the first and second terms of school, as well as in our higher education.

• In exercise 13.2 Class 9 Maths, The formulas for calculating surface areas and volume for the right circular cylinder are well explained and these formulas will help in solving the competitive questions, hence we must make a note of those formulas.

## Key Features of 9th Class Maths Exercise 13.2 Answers

1. Step-by-Step Solutions: The NCERT Solutions fo class 9 ex 13.2 provide detailed, step-by-step explanations for each problem, making it easier for students to grasp the concepts.

2. Practical Application: This exercise presents practical problems that require students to calculate surface areas and volumes of various 3D shapes, honing their problem-solving skills.

3. Clear and Understandable Language: Class 9 maths ex 13.2 Solutions are presented in clear and understandable language, ensuring that students can comprehend the concepts with ease.

4. PDF Format: Students can access the ex 13.2 class 9 solutions in PDF format, enabling them to download and use the materials offline, making learning convenient and accessible.

5. Foundation for Advanced Geometry: Mastery of surface areas and volumes of 3D shapes in this exercise prepares students for more complex geometry topics they will study in higher classes.

Also See:

## NCERT Solutions of Class 10 Subject Wise

1. Define the right circular cylinder, according to NCERT solutions for Class 9 Maths chapter 13 exercise 13.2 .

Solution :

A right circular cylinder has a closed circular surface with two parallel bases on both ends and elements that are perpendicular to its base, according to NCERT solutions for Class 9 Maths chapter 13 exercise 13.2.

2. How many parts are there in the right circular cylinder and what are they?

Solution :

The right circular cylinder is divided into three parts. They are

• Top circular base

• Curved lateral face

• Bottom circular face

3. What are the types of surface area?

Solution :

There are two types of surface area . They are

• Lateral Surface Area

• Total Surface Area

4. Define the lateral surface area of the right circular cylinder.

Solution :

The area covered by the curved surface of the cylinder is known as the curved surface area of the right circular cylinder which is often referred as lateral surface area of the right circular cylinder.

5. The lateral surface area of the right circular cylinder is ________

Solution :

The lateral surface area of the right circular cylinder is  2πrh

6. The total surface area of the right circular cylinder is ________

Solution :

The right circular cylinder has a total surface area of 2πr(h+r)

7. The surface areas are calculated in _______ units

Solution :

The surface areas are measured in terms of square units .

8. A closed right circular cylinder's total surface area is equal to the sum of ________

Solution :

The area of the lateral surface plus the area of the two bases equals the total surface area of a closed right circular cylinder.

9. The number of surfaces in the right cylinder is ______
1. 4

2. 3

3. 2

4. 1

Solution :

The number of surfaces in the right cylinder is 3 . Option (b) 3

10. According to NCERT solutions for Class 9 Maths chapter 13 exercise 13.2 ,How is the total surface area calculated from the lateral surface area and area of the circle?

Solution :

The total surface area of a closed right circular cylinder is computed by adding the area of the lateral surface and the area of the two bases, according to NCERT solutions for Class 9 Maths chapter 13 exercise 13.2 .

TSA=CAS+2(Area of a circle)

= 2πrh+2π(r^2)

TSA=2πr(h+r)

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